Laws of MotionMedium
Question
A block of mass m = 2 kg is resting on a rough inclined plane of inclination 30o as show in figure. The coefficient of friction between the block and the plane on the block, so that does slip on the plane. (g = 10 m/s2)
Options
A.zero
B.6.24N
C.2.68N
D.4.34N
Solution
N = F + mgcosθ, f = mgsinθ but f ≤ μN so mgsinθ ≤ (F + mgcosθ)
⇒ F ≥ mg
⇒ Fmin = 2 × 10
= 20(1 - 0.866) = 2.68 N
⇒ F ≥ mg
⇒ Fmin = 2 × 10
= 20(1 - 0.866) = 2.68 N
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